Answer to Question #278624 in Differential Equations for Ravi

Question

(x^2D^2-xD+1)y=sin (logx)

Answer 1

$Given\mathrm{:}\ \ \ \left(x^{\mathrm{2}}D^{\mathrm{2}}\ \ \ -\ \ xD\ \ +\ \ \mathrm{1}\right)y\ \ =\ \ \mathrm{sin}\left(\mathrm{log}\left(x\right)\right) \\ Let\ \ z=\mathrm{log}\left(x\right)\ \ ,\ \ such\ \ that\ \ x=e^z \\ \\ D\left(D-\mathrm{1}\right)y-Dy\ \ +\ \ y\ \ =\ \ \mathrm{sin}\left(z\right) \\ \\ \left(D^{\mathrm{2}}-\mathrm{2}D+\mathrm{1}\right)y\ \ =\ \ \mathrm{sin}\left(z\right) \\ \\ A.E.\ \ is\ \ \ \ m^{\mathrm{2}}-\mathrm{2}m{}{}\ +\ \mathrm{1}=0\ \ \ \\ \\ \Rightarrow \ \ \left(m-\mathrm{1}\right)\left(m-\mathrm{1}\right)=0\ \ \ ,\ \ m=\mathrm{1,1}\ \ \ \ \ \left(repeated\ \ root\right) \\ \\ \Rightarrow y_c=\left(A+Bz\right)e^z\ \ ...................\left(\mathrm{1}\right) \\ \\ Let\ \ y_p=\frac{\mathrm{1}}{D^{\mathrm{2}}-\mathrm{2}D+\mathrm{1}}\mathrm{sin}\left(z\right) \\ \\ y_p=\frac{\mathrm{1}}{-{\mathrm{1}}^{\mathrm{2}}-\mathrm{2}D+\mathrm{1}}\mathrm{sin}\left(z\right)\ \ =\frac{\mathrm{1}}{-\mathrm{2}D}\mathrm{sin}\left(z\right)\ \ =\frac{\mathrm{cos}\left(z\right)}{\mathrm{2}} \\ \\ y_p=\frac{\mathrm{cos}\left(z\right)}{\mathrm{2}}..........................\left(\mathrm{2}\right) \\ \\ y=y_c+y_p \\ \\ y=\left(A+Bz\right)e^z\ +\frac{\mathrm{cos}\left(z\right)}{\mathrm{2}} \\ \\ y=\left(A+B\mathrm{log}\left(x\right)\right)x\ +\frac{\mathrm{cos}\left(\mathrm{log}\left(x\right)\right)}{\mathrm{2}}$